2. solutia initiala contine atat CH3COOH cat si H2O ⇒
⇒ n1moli CH3COOH + Na ⇒ CH3COONa + 1/2 H2
n2 moli H2O + Na ⇒ NaOH + 1/2H2
60n1 + 18n2 = 110g 30n1 + 9n2 = 55
n1 + n2 = 122,6/23
21n1 = 3 n1 = 1/3 moli CH3COOH (20g) n2 = 5moli H2O (90g)
ms = 110g md = 20g c = 18,18%
3. solutia de CH3COOH contine acid : n a = 200·0,2/60 = 2/3 moli
solutia de KOH contine baza : n b = 0,1ms/56
a) la neutralizare : n a = n b 2/3 = 0,1ms/56 0,3ms = 112
ms KOH = 1120/3 = 373,33 g
n acetat = n KOH = 2/3 m sare = 2/3 ·98 = 65,33 g
